Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 350 1 c Solution Created 2026-10-03 Updated 2026-10-05
Use the Hellinger distance convention in the original PDF:The TeX transcription incorrectly places the integral outside the outer square root in the definition; the displayed formula for the squared distance and the original PDF both use the convention above. Although Lebesgue measure is not a probability measure, we may use it as the common dominating measure: the integral is unchanged by replacing it with any equivalent dominating probability measure.
For unit-variance normal distributions, put and . The product of the square roots of their probability density functions isThe final factor integrates to one, soFor the bound at arbitrary , use the given formula for the Hellinger distance between normal distributions and writeSince for ,Therefore one valid choice is . In particular this bound is uniform when the two standard deviations stay bounded away from zero.